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Living_dreams
find all local max,min and/or saddle point of f(x,y)=4+x^3+y^-3xy
would you plz write the function using equation
hmm, something to do with:\[f_{xy}f_{yx}-(f_{xx})^2\] or did i recall that incorrectly?
almost had it :) \[f_{xx}f_{yy}-(f_{xy})^2\]
\[4+x ^{3}+y ^{3}-3xy\]
it fxx.fyy-(fxy)^2
i jst want to verify the answer for it !
what answer did you get?
\[fx = 2x-3y \] \[fy= 2y-3x\] on solving both equations the point is (0,0) \[fxx = 2\] , \[fyy= 2\] \[fxy = \] \[D = -5 <0 saddle point\]
that's not quite right I don't think
\[f(x,y)=4+x^3+y^3-3xy\]\[f_x=3x^2-3y=0\implies y=x^2\]\[f_y=3y^2-3x=3x^4-3x=3x^3(x-1)=0\]\[ x=\{0,1\},y=\{0,1\}\]
its x^3 so i have (0,0) ,(1,1)and a (-1,1) i am not sure if (-1,1) should be there oor not
@Living_dreams I solved what you mentioned in your comment
there is no x=-1 solution, I don't see how you get that
@Avva I don't see how you get your fx and fy
Ooh I saw the powers on X ^ y 2 not 3
@Avva thanks for ur help but i was wonder abt the max and min points! n\ @TuringTest : 3y^2-3x=0 i sub y=x^2 from other equation so it will be 3x(x^3-1)=0 therefore you have x =0,1,-1 i am not sure abt -1 though !! tht was my question
x^3=1 does not have x=-1 as a solution since (-1)^3=-1, not 1
I did screw up my factoring earlier though, you were right about that :P
3x(x^3-1)=0 3x=0 -> x=0 x^3=1 -> x=1
@TuringTest that alrite!! :) so now my question is x^3=1 so will i hv x=-1 if i solve for x ?
no, again, because \((-1)^3\neq1\) so \(x^3=1\) has only one real solution, x=1 x=-1 is NOT a solution
the other two roots are complex not real
@TuringTest oh ok thanks a lot for ur help!! so the final ans will be (0,0 ) is a saddle point and (1,1) will be a local minimum !
@Avva :thanks for ur help !! :)
@Living_dreams URW :) sorry I didn't get that the powers are 3 at first
it okay !!i knw it get hard to tell the difference between 2 and 3 if we type it in equation form !!